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Sinusoidal Graph - sub intervals |
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| Mar16-11, 02:13 AM | #1 |
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Sinusoidal Graph - sub intervals
1. The problem statement, all variables and given/known data
I can find the graphs amp and period. The only problem is finding the sub points or sub intervals. Say... Y = 3 sin (4x) Amp = 3 Period = 2pi/4 = pi/2 But.. don't know how to get the key points of the sub interval. The text book says I have to divide interval [0, pi/2] into four sub intervals Each of length pi/2 divided by 4. Then they got (0,0), (pi/8, 3), (pi/4, 0), (3pi/8, -3), (pi/2, 0) I don't understand how they got these. Thanks 2. Relevant equations 3. The attempt at a solution |
| Mar16-11, 02:38 AM | #2 |
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Recognitions:
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Draw the base graph y=sin(x) with [itex]0\leq x\leq 2\pi[/itex], It has a value of 0 at [itex]x=0,\pi,2\pi[/itex] and a value of 1 and -1 at [itex]x=\pi/2, 3\pi/2[/itex] respectively.
Basically, every sine graph of the form [itex]y=Asin(Bx)[/itex] will still have this same shape, but the amplitude (A) and period (B) will be different from the base graph y=sin(x). What you should take away from this is that if the period of sin(x) is [itex]2\pi[/itex], then in between the two ends of the period 0 and [itex]2\pi[/itex] which is [itex]\pi[/itex], it will also be 0, and in between 0 and its half way mark which is [itex]\pi[/itex] we get the value of its amplitude (in this case 1), and between the half way mark and the end, [itex]\pi[/itex] and [itex]2\pi[/itex] we get the negative of its amplitude, -1. |
| Mar16-11, 03:30 AM | #3 |
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I know if period = 2 pi then the middle point would me pi, but what if it has a different period. I don't know what to label on the graph on the middle part. |
| Mar16-11, 03:51 AM | #4 |
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Recognitions:
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Sinusoidal Graph - sub intervals
If the period is [itex]2\pi[/itex] then middle is half of that [tex]\frac{2\pi}{2}=\pi[/tex]. If the period is some number x then the middle is x/2.
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| Mar16-11, 04:23 AM | #5 |
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Ahhh I see now. Thank you my friend.
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| Mar16-11, 04:55 AM | #6 |
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Recognitions:
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Good luck!
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